Problem: Graph this system of equations and solve. $-6x+2y = 6$ $-2x-4y = 16$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $9$ $10$ $\llap{-}2$ $\llap{-}3$ $\llap{-}4$ $\llap{-}5$ $\llap{-}6$ $\llap{-}7$ $\llap{-}8$ $\llap{-}9$ $\llap{-}10$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $9$ $10$ $\llap{-}2$ $\llap{-}3$ $\llap{-}4$ $\llap{-}5$ $\llap{-}6$ $\llap{-}7$ $\llap{-}8$ $\llap{-}9$ $\llap{-}10$ Click and drag the points to move the lines.
Convert the first equation, $-6x+2y = 6$ , to slope-intercept form. $y = 3 x + 3$ The y-intercept for the first equation is $3$ , so the first line must pass through the point $(0, 3)$ The slope for the first equation is $3$ . Remember that the slope tells you rise over run. So in this case for every $3$ positions you move up $1$ position to the right. $3$ positions up from $(0, 3)$ is $(1, 6)$ Graph the blue line so it passes through $(0, 3)$ and $(1, 6)$ Convert the second equation, $-2x-4y = 16$ , to slope-intercept form. $y = -\dfrac{1}{2} x - 4$ The y-intercept for the second equation is $-4$ , so the second line must pass through the point $(0, -4)$ The slope for the second equation is $-\dfrac{1}{2}$ . Remember that the slope tells you rise over run. So in this case for every $1$ position you move down (because it's negative) You must also move $2$ position to the right. $2$ positions to the right. Graph the green line so it passes through $(0, -4)$ and $(2, -5)$ The solution is the point where the two lines intersect. The lines intersect at $(-2, -3)$.